A solid usually expands when heated and contracts when cooled. If the solid is free to change length, no thermal
stress is produced. If the solid is constrained, prevented expansion or contraction creates mechanical strain and
stress.
1. Free thermal strain
Linear thermal strain is
\[
\boxed{
\varepsilon_{\mathrm{th}}=\alpha\Delta T
}.
\]
Here, \(\alpha\) is the coefficient of linear thermal expansion and \(\Delta T\) is the temperature change.
2. Free thermal expansion
If a rod of original length \(L\) is free to expand, its change in length is
\[
\boxed{
\Delta L_{\mathrm{free}}=\alpha L\Delta T
}.
\]
Heating gives positive \(\Delta L_{\mathrm{free}}\), while cooling gives negative \(\Delta L_{\mathrm{free}}\).
3. Free expansion produces no stress
If the rod is free, the total strain equals the thermal strain:
\[
\varepsilon_{\mathrm{total}}=\varepsilon_{\mathrm{th}}.
\]
Therefore the mechanical strain is zero:
\[
\varepsilon_{\mathrm{mech}}=0.
\]
So
\[
\sigma=E\varepsilon_{\mathrm{mech}}=0.
\]
4. Fully fixed rod
If a rod is fixed between rigid supports, its total change in length is zero:
\[
\varepsilon_{\mathrm{total}}=0.
\]
Total strain is the sum of thermal strain and mechanical strain:
\[
\varepsilon_{\mathrm{total}}
=
\varepsilon_{\mathrm{th}}
+
\varepsilon_{\mathrm{mech}}.
\]
Therefore,
\[
\varepsilon_{\mathrm{mech}}
=
-\varepsilon_{\mathrm{th}}
=
-\alpha\Delta T.
\]
The thermal stress is
\[
\boxed{
\sigma=-E\alpha\Delta T
}.
\]
With the usual sign convention, heating a fixed rod gives compressive stress, and cooling a fixed rod gives
tensile stress.
5. Partially constrained rod
If only a fraction \(c\) of the free expansion is prevented, then
\[
\varepsilon_{\mathrm{mech}}=-c\alpha\Delta T,
\]
and
\[
\boxed{
\sigma=-cE\alpha\Delta T
}.
\]
The value \(c=0\) means free expansion, and \(c=1\) means fully fixed ends.
6. Gap before contact
Sometimes a rod has a small expansion gap before it contacts a support. If the free expansion is smaller than the
gap, no stress develops:
\[
\Delta L_{\mathrm{free}}\le g_{\mathrm{ap}}
\quad\Rightarrow\quad
\sigma=0.
\]
If the free expansion exceeds the gap, the remaining expansion is prevented. Then
\[
\varepsilon_{\mathrm{total}}=\frac{g_{\mathrm{ap}}}{L}.
\]
The mechanical strain is
\[
\varepsilon_{\mathrm{mech}}
=
\frac{g_{\mathrm{ap}}}{L}
-
\alpha\Delta T.
\]
Therefore,
\[
\boxed{
\sigma=
E\left(\frac{g_{\mathrm{ap}}}{L}-\alpha\Delta T\right)
}.
\]
7. Support reaction force
If the rod has cross-sectional area \(A\), the axial support reaction is
\[
\boxed{
R=\sigma A
}.
\]
The sign indicates tensile or compressive direction, while the magnitude gives the reaction force size.
8. Safety check
A safe design compares the thermal stress with an allowable stress:
\[
\sigma_{\mathrm{allow}}=\frac{\sigma_{\mathrm{limit}}}{n},
\]
where \(n\) is the required safety factor.
The design is acceptable if
\[
|\sigma|\le\sigma_{\mathrm{allow}}.
\]
9. Maximum safe temperature change
For a fully fixed rod, the stress magnitude is
\[
|\sigma|=E\alpha|\Delta T|.
\]
So the maximum safe temperature change is approximately
\[
\boxed{
|\Delta T|_{\max}
=
\frac{\sigma_{\mathrm{allow}}}{E\alpha}
}.
\]
10. Typical thermal expansion coefficients
| Material |
Typical \(\alpha\) |
Typical Young’s modulus |
| Steel |
\(11{-}13\times10^{-6}/^\circ\mathrm{C}\) |
\(\approx200\ \mathrm{GPa}\) |
| Aluminum |
\(22{-}24\times10^{-6}/^\circ\mathrm{C}\) |
\(\approx69\ \mathrm{GPa}\) |
| Copper |
\(16{-}17\times10^{-6}/^\circ\mathrm{C}\) |
\(\approx110\ \mathrm{GPa}\) |
| Concrete |
\(10{-}12\times10^{-6}/^\circ\mathrm{C}\) |
\(\approx20{-}40\ \mathrm{GPa}\) |
| Glass |
\(8{-}9\times10^{-6}/^\circ\mathrm{C}\) |
\(\approx50{-}90\ \mathrm{GPa}\) |
11. Worked example
A steel rod has
\[
\alpha=12\times10^{-6}/^\circ\mathrm{C},
\qquad
E=200\ \mathrm{GPa},
\qquad
\Delta T=80^\circ\mathrm{C}.
\]
The thermal strain is
\[
\varepsilon_{\mathrm{th}}=\alpha\Delta T.
\]
\[
\varepsilon_{\mathrm{th}}
=
(12\times10^{-6})(80)
=
9.60\times10^{-4}.
\]
If the rod is fully fixed, the mechanical strain must cancel the thermal strain:
\[
\varepsilon_{\mathrm{mech}}=-9.60\times10^{-4}.
\]
Therefore,
\[
\sigma=E\varepsilon_{\mathrm{mech}}.
\]
\[
\sigma
=
(200\times10^9)(-9.60\times10^{-4})
=
-1.92\times10^8\ \mathrm{Pa}.
\]
\[
\boxed{
\sigma=-192\ \mathrm{MPa}
}.
\]
The negative sign means the stress is compressive.
Key idea: free thermal expansion creates strain but no stress. Thermal stress appears when expansion or contraction is prevented.
